Unit 9
Practice
Test – SOLUTIONS (KEY)
Kinetics
Part I: Multiple Choice Questions
Questions 1–2
2 NO(g) + H2(g) N2(g) + 2 H2O(g)
The oxidation of diatomic hydrogen by nitrogen oxide occurs at 1300 ˚C according to the reaction shown above. The experimental rate law of the reaction is: r = k[NO]2
[H2]
1. What is the order of the reaction with respect to NO ?
(A) 1 (B) 2 (C) 3 (D) 5 E) 6
The order of the reaction with respect to a species that appears in the rate law is the exponent on the concentration term for that species. Here the exponent for [NO] is 2 and the reaction would be second order in [NO]. The correct choice was
B.
2. According to the rate law for the reaction, an increase in the concentration of diatomic hydrogen has what effect on this reaction?
(A) The rate of reaction increases. (B) The rate of reaction decreases. (C) The rate constant increases (D) The rate constant decreases
(E) Neither the rate nor the rate constant is changed.
The rate of reaction can depend on the concentration of species, temperature, contact area between reactants or addition of a catalyst. The rate law directly shows the dependence of the rate of reaction on concentration. The other dependencies are buried in the Arrhenius rate constant. If r = k[NO]2
[H2], the addition of diatomic hydrogen will increase the rate of reaction. The correct choice was A.
3. A(g) + 2 B(g) C(g)
When the concentration of substance A in the reaction above is halved, all other factors being held constant, it is found that the rate of the reaction remains unchanged. The most probable explanation for this observation is that (A) the order of the reaction with respect to substance A is 1/2
(B) substance A is not involved in any of the steps in the mechanism of the reaction
(C) substance A is not involved in the rate–determining step of the mechanism, but is involved in subsequent steps (D) substance A is probably a catalyst, and as such, its effect on the rate of the reaction does not depend on its
concentration
(E) the reactant with the largest coefficient in the balanced equation generally is rate determining
The rate of reaction can depend on the concentration of species, temperature, contact area between reactants or addition of a catalyst. The rate law directly shows the dependence of the rate of reaction on concentration. The other dependencies are buried in the Arrhenius rate constant. The fact that the concentration of a reactant does not affect the rate of reaction means that the rate law is zero order with respect to the species but it does not mean that the species doesn’t participate in the reaction. It just means that the species doesn’t participate in the step of the reaction that is rate determining. Choice C
4. Step 1. Cl2(g) 2Cl(g) (fast equilibrium) Step 2. Cl(g) + CHCl3(g) HCl(g) + CCl3(g)(slow)
Step 3. Cl(g) + CCl3(g) CCl4(g) (fast)
Chlorine gas reacts with chloroform (CHCl3) to yield hydrogen chloride and carbon tetrachloride. This
decomposition is believed to occur according to the reaction mechanism above. The rate law that is consistent with this mechanism is given by which of the following?
(A) Rate = k [Cl][CHCl3]
(B) Rate = k [Cl2][CHCl3]
(C) Rate = k [Cl2]2[CHCl3] (D) Rate = k [Cl2]1/2[CHCl3]
(E) The rate mechanism is not consistent with the stated reaction
Step 1: Check to make sure that the steps of the proposed reaction mechanism add up to the overall reaction: Overall: Cl2(g) + Cl(g) + CHCl3(g) + Cl(g) + CCl3(g) 2Cl(g) + HCl(g) + CCl3(g) + CCl4(g)
Cancel terms and you get: Cl2(g) + CHCl3(g) 2 HCl(g) + CCl4(g) The rate mechanism is consistent with the stated reaction.
Step 2: generate a rate law from the slow step using mass-action kinetics. r = k[Cl][CHCl3]
The rate law from the rate determining step (slow step) contains a species that does not appear in the overall reaction, Cl(g). Use the prior fast step and the condition of equilibrium to eliminate it.
K1 = solve for the [Cl] = K’1[Cl2]1/2 and substitute this expression into the rate law from the slow step to get: r = k K’1[Cl2]1/2 [CHCl3] or r = k’[Cl2]1/2 [CHCl3]
The rate law indicates that the reaction kinetics are one-half order with respect to [Cl2] and first order with respect to [CHCl3]. The correct choice was D.
5. Which of the following is a graph that describes the pathway of reaction that is exothermic and has the highest acti-vation energy? A. B. Po te nt ia l E ne rg y
Reaction Coordinate Po
te nt ia l E ne rg y Reaction Coordinate
C. D. E.
Po te nt ia l E ne rg y
Reaction Coordinate Po
te nt ia l E ne rg y
Reaction Coordinate Po
te nt ia l E ne rg y Reaction Coordinate
6. rate = k[X]2
For the reaction whose rate law is given above, a plot of which of the following is a straight line? A. [X] versus time B. ln [X] versus time C. 1/[X] versus time
D. [X] versus 1/time E. ln [X] versus 1/time
In order to determine the order of the reaction when given concentration versus time data, you need to make the graphs associated with the integrated rate laws for zero, first and second order. The linear relationships (integrated rate laws) are as follows:
Zero order reaction: [A] = -kt + [A]o. Plot [A] vs. t
First order reaction: ln[A] = -kt + ln[A]o. Plot ln[A] vs. t
Second order reaction: 1/[A] = kt + 1/[A]o. Plot 1/[A] vs. t
The plot that yields a straight line is the one that indicates the order of the kinetics. Here since the reaction is second order, a plot of 1/[X] vs. t would be linear. The correct choice was C.
7. (CH3)3CCl(aq) + OH- (CH3)3COH(aq) +
Cl-For the reaction represented above, the experimental rate law is given as follows. Rate = k[(CH3)3CCl]
If some solid sodium hydroxide is added to a solution that is 0.010–molar in (CH3)3CCl and 0.10–molar in NaOH, which of the following is true? (Assume the temperature and volume remain constant.)
A. Both the reaction rate and k increase. B. Both the reaction rate and k decrease.
C. Both the reaction rate and k remain the same. D. The reaction rate increases but k remains the same. E. The reaction rate decreases but k remains the same.
The rate law indicates that the rate determining step is not dependent on the concentration of hydroxide. Increasing the concentration of hydroxide will therefore have no affect on the rate of reaction. The rate constant does not have a dependency on concentration. It does however have a dependency on temperature but the temperature remains constant and therefore k also remains constant. The correct choice was C.
8. Relatively fast rates of chemical reaction are associated with which of the following? A. The presence of an inhibitor
B. Low temperature
C. Low concentration of reactants D. Strong bonds in reactant molecules E. Low activation energy
9. The energy diagram for the reaction X + Y Z is shown above. The addition of a catalyst to this reaction would not cause a change in which of the indicated energy differences?
A. I only B. II only
C. III only D. I and II only
E. I, II, and III
The addition of a catalyst causes a change in the height of the activation energy barrier. Here that corresponds to I for the forward reaction and II for the reverse reaction. The heat of reaction, i.e. the energy difference between the reactants and products, is not affected by the addition of a catalyst. The correct choice was C.
10. The reaction of nitric oxide with hydrogen occurs according to the reaction: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
The rate constant for this reaction is, k = 1.1 M-2s-1. What is the overall order of the rate law?
(A) 0 (B) 1 (C) 2 (D) 3 E) 4
Understand that you can determine the order of a rate law from the units of the rate constant. The rate of reaction is typically in terms of M•s-1 (that’s the change in molar concentration per second). Depending on the order of the
concentration terms, the units for the rate constant will change in order to provide a product of terms with the units M•s-1. So, for zero order kinetics, the units of k are M•s-1. For first order kinetics (sum of the exponents on the concentration terms = 1) the units of k are s-1. For second order kinetics (sum of the exponents on the concentration terms = 2) the units of k are M-1•s-1. For third order kinetics (sum of the exponents on the concentration terms = 3) the units of k are M-2•s-1. Here the units of the rate constant are consistent with third order kinetics overall. The correct choice was D.
11. The reaction of peroxydisulfate ion (S2O82-) with iodide ion (I-) is
S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
The initial rate of formation of sulfate was determined to be 2.1 x 10-4 M/s. What is the initial rate of disappearance of iodide?
(A) -2.1 x 10-4 M/s (B) -3.15 x 10-4 M/s (C) 1.1 x 10-4 M/s (D) 2.1 x 10-4 M/s (E) 3.15 x 10-4 M/s
Two things: the rate of appearance/disappearance of the species in a chemical reaction is related by the Stoichiometry of the reaction. Here that means the rate of disappearance of iodide will be 3/2 the rate of appearance of sulfate. The second thing is that rates of reaction are never negative. By saying it is the rate of disappearance tells you the concentration is decreasing. You don’t need to report it as a negative number. 3/2(2.1 x 10-4 M/s) = 3.15 x 10-4 M/s. The correct choice was E.
[Reactant]
(mol/L) 0.02000 0.01574 0.01298 0.01105 0.00962
12. A reaction was observed for 200 seconds and the concentration the reactant was monitored as a function of time. Which of the following best describes the order and half-life of the reaction?
Reaction Order Half-life (seconds)
(A) First 100
(B) First 200
(C) Second 100
(D) Second 150
(E) Second 200
The half life part was easy. Just look at the data and you can see that very close to half of the initial concentration is used up at 200 seconds. That rules out every choice except (B) and (E). Next, check the data to see if it is consistent with first or second order kinetics.
First order reaction: ln[A] = -kt + ln[A]o. Plot ln[A] vs. t
Second order reaction: 1/[A] = kt + 1/[A]o. Plot 1/[A] vs. t
The plot that yields a straight line is the one that indicates the order of the kinetics. Test a couple of numbers and you’ll see that the data are linear for second order kinetics. The correct choice was E.
14. The Arrhenius form of the rate constant is typically used in rate laws. The value of the rate constant is dependent on: I. The concentration of reactants
II. Temperature
III. The presence of a catalyst
(A) I, II, and III (B) I and II (C) II only
(D) II and III (E) The rate constant is a constant and does not change
The Arrhenius form of the rate constant is . The dependency on activation energy and temperature are shown in
Part II Free Response
:1. NH4+ (aq) + NO2- (aq) N2 (g) + 2 H2O (l)
The following data were obtained at 25 ˚C.
Initial Rate of Reaction,
(mol.L-1.sec-1) Initial [NH4+]o, (mol.L-1)
Initial [NO2-]o, (mol.L-1)
10.8 x 10-7 0.0200 0.200
10.8 x 10-7 0.200 0.0202
21.6 x 10-7 0.200 0.0404
21.5 x 10-7 0.0400 0.200
(a) Give the rate law for this reaction from the data above.
The data is concentration versus rate. When presented with concentration versus rate data, you want to set up a ratio of rate laws for different experimental runs in order to solve for the exponents in the rate law. Typically, as is the case here, some of the runs will have used the same concentration values so that the concentration dependence of that species drops out of the ratio. Take a ratio of run 3 over run 2, the dependence on the concentration of NH4+ drops out (becomes 1), the rate constants cancel and you are left with (0.0404/0.0202)M = (21.6 x 10-7/10.8 x 10-7). Reduce the numbers and you get (2)M = (2). M = 1 and the rate law is therefore first order in [NO2-]. Now use runs 4 and 1 to determine the rate law
dependence on the concentration of NH4+. The concentration of NO2- drops out (becomes 1), the rate constants cancel and you are left with (0.0400/0.0200)N = (21.5 x 10-7/10.8 x 10-7). N = 1 and the rate law is therefore first order in [NH4+]. The rate law is r = k[NH4+][NO2-]
(b) Calculate the specific rate constant for this reaction and specify its units.
Once you know the rate law and have concentration versus rate data, you can use any one of the runs to solve for the rate constant. r = k[NH4+][NO2-]. Using run 1: 10.8 x 10-7 M•s-1 = k (0.0200)(0.200). Solve for k and you get 2.7 x 10-4 M-1•s-1.
(c) The reaction is allowed to proceed until a volume of N2 (g) equal to 250 mL is produced. The initial reaction
concentrations in a 1.0 L container are [NH4+]o = 0.60 molar, [NO2-]o = 0.60 molar and the N2 (g) is collected over water at 1 atm of pressure. The vapor pressure of water at 25 ˚C = 23.4 mm Hg. What is the rate of reaction at the moment when the 250 mL container is filled. (7 points)
First determine the moles of N2 (g) produced. Use the Ideal Gas Law.
Ptot = PN2 + PH2O; PN2 = 760 mm Hg – 23.4 mm Hg = 736.6 mm Hg x (1atm/760 mm Hg) = 0.9692 atm
PV= nRT; n=
(
0.9692atm)
(0.250L)(
0.0821 L∙ atmmol ∙ K
)
(298K)=0.0099moles
ICE table, all in 1.0 L, so just use molar numbers as they are x = 0.0099, so 0.60 M – 0.0099 M = 0.59 M NH4+ (aq) NO2- (aq) N2 (g) 2 H2O(l)
I 0.60 0.60 0 0
C -x -x +x +2x
r = k[NH4+][NO2-] = 2.7 x 10-4(0.59)(0.59) rate = 9.4 x 10-5
2. CH3COOC2H5 (aq) + OH- (aq) CH3COO- (aq) + 2 C2H5 OH (aq)
The rate of the above reaction was measured at a number of temperatures and the following values of the rate constant were determined:
T = 298 K k = 0.101 M-1•s-1 T = 318 K k = 0.332 M-1•s-1
What is the activation energy associated with this reaction? (9 points)
Starting with the Arrhenius form of the rate constant: this relationship can be linearized by taking the natural
log of both sides of the equation: . At two temperatures, the equations can be subtracted from each other to yield:
. Plug in the numbers and solve for the activation energy.
EA = 46.9 kJ.
3. The following data were collected for the gas phase decomposition of nitrogen dioxide:
NO2 (g) NO(g) + O2 (g)
Time (seconds) 0 5 10 15 20
[NO2] (mol/L) 0.2000 0.034 0.018 0.0124 0.0094
Plot this data on the appropriate coordinate system to determine the reaction order with respect to [NO2]. (14 points)
Zero order reaction: [A] = -kt + [A]o. Plot [A] vs. t
First order reaction: ln[A] = -kt + ln[A]o. Plot ln[A] vs. t
Second order reaction: 1/[A] = kt + 1/[A]o. Plot 1/[A] vs. t
Time (s) [NO2] ln[NO2] 1/[NO2]
0 0.2000 -1.6 5.0
5 0.034 -3.4 29.4
10 0.018 -4.0 55.6
15 0.0124 -4.4 80.6
Zero order:
1 2 3 4 5
0 0.05 0.1 0.15 0.2 0.25
time
[N
O
2
]
First order:
1 2 3 4 5
-5.0 -4.5 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0
time
ln
[N
O
2
]
Second order:
1 2 3 4 5
0.0 20.0 40.0 60.0 80.0 100.0 120.0
time
1
/[
N
O
2
]
Any time you are presented data that relates concentration to time, you want to use the integrated version of the rate law. For a second order reaction that is: 1/[A] = kt + 1/[A]o. The only unknown in the integrated rate law is the rate constant.
